Suppose that $y^3$ varies inversely with $\sqrt[3]{z}$. If $y=2$ when $z=1$, find the value of $z$ when $y=4$. Express your answer in simplest fractional form.
Solution: Since $y^3$ varies inversely with $\sqrt[3]{z}$, $y^3\cdot\sqrt[3]{z}=k$ for some constant $k$. If $y=2$ when $z=1$, then $k=2^3\cdot\sqrt[3]{1}=8\cdot1=8$. Thus, when $y=4,$ we have: \begin{align*} (4)^3\sqrt[3]{z}& =8
\\ 64\sqrt[3]{z}&=8
\\\Rightarrow\qquad \sqrt[3]{z}&=\frac18
\\\Rightarrow\qquad z&=\left(\frac18\right)^3
\\ z&=\boxed{\frac1{512}}
\end{align*}